Why does the fact that $f^{'}(t)$ being continuous insure that $\int_{0}^{T}f^{'}(t)^2 dt$ is finite

Question

Why does the fact that $f^{'}(t)$ being continuous insure that $\int_{0}^{T}f^{'}(t)^2 dt$ is finite

It might be obvious, but I can't see why when we know that that $f^{'}(t)$ is continuous we can conclude that $\int_{0}^{T}f^{'}(t)^2 dt$ is finite.

Any help, reference to some result/course is appreciated thank you

Answer 1

If $f'$ is continuous on $[0,T]$ then $(f')^2$ is also continuous there (product of two continuous functions). By Weierstrass' extreme value theorem $(f')^2$ admits then a maximum $M$ and a minimum $m$ on $[0,T]$, i.e. it is bounded. This implies that $$\left| \int_0^T f'(t)^2 \, dt \right| \le \int_0^T |f'(t)|^2 \, dt \le M \cdot T < \infty.$$

Answer 2

Because any continuous function on a compact set says $[0,T]$ is bounded,

Therefore for some $m>0$ $$|f'(t) |\le m,~~~\forall t\in [0,T], ~~~$$

thus, $$\int_0^Tf'(t) ^2dt\le Tm^2~$$